$\overline{BC}=10$ $\overline{AB} = {?}$ $A$ $C$ $B$ $?$ $10$ $ \sin( \angle ABC ) = \frac{3\sqrt{109} }{109}, \cos( \angle ABC ) = \frac{10\sqrt{109} }{109}, \tan( \angle ABC ) = \dfrac{3}{10}$
Solution: $\overline{BC}$ is adjacent to $\angle ABC$ $\overline{AB}$ is the hypotenuse (note that it is opposite the right angle) SOH CAH TOA We know the adjacent side and need to solve for the hypotenuse so we can use the cos function (CAH) $ \cos( \angle ABC ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\overline{BC}}{\overline{AB}} = \frac{10}{\overline{AB}} $ $ \overline{AB}=\frac{10}{\cos( \angle ABC )} = \frac{10}{\frac{10\sqrt{109} }{109}} = \sqrt{109}$